Module equivalence

There is a problem in a book I'm not quite understanding.
Let M be an R-module and let I=Ann(M). Show that M can be regarder as an R/I-Module where scalar multiplication is given by the rule m(I+r)=mr

I don't understand what they mean by "regarded as". Am I suppose to show there is an isomorphism? (I don't know how to do that between modules over different rings), or just a bijection? Should I create a one-to-one mapping from the scalars R to R/I instead?

Really, they probably just want you to show that the underlying additive group of M, together with that scalar multiplication, is an R/I-module.

Bonus points for explicitly showing that, if [itex]\pi[/itex] is the projection from R to R/I, that [itex]r \cdot m = \pi(r) \cdot m[/itex]. (where the two [itex]\cdot[/itex] symbols refer to the appropriate scalar products)

One main point is that, in the expression [itex]r \cdot m[/itex] where r is in R and m is in M, it doesn't matter if we mean this product as the scalar product on M, or if we are using r to name an element of R/I, and the product is the scalar product on the R/I-module defined in the problem; the result of the arithmetic expression is the same.

Ok, thank you Hurkyl!
Just out of curiosity, what exactly does this result mean, the fact that the set M can be regarded as both an R-module and an R/I-module? Is this mostly a consequence of the properties of the quotient ring R/I (for example of, like you pointed out, the canonical homomorphism from R to R/I)?
I'm finding it hard to wrap my head around it because R and R/I are not isomorphic (possibly not even of the same order if finite) yet behave the same as scalars over M?

The more opportunities you have to use this fact, the more it will make sense.

You already know that all of the elements of I behave exactly the same as scalar multipliers on M. So what do you get when you impose that they are all congruent?

Incidentally, there are two well-behaved constructions related to a ring homomorphism f:R --> S.

(The underlying group of) any S-module M can be given an R-module structure defined by [itex]r\cdot m = f(r)\cdot m[/itex].

By the above, S itself can be viewed as an R-module. For any R-module M, the (underlying group of) the tensor product [itex]S \otimes_R M[/itex] can be given an S-module structure defined by [itex]s \cdot (s' \otimes m) = (ss') \otimes m[/itex].

When S is the quotient ring R/I, the constructions are even more well-behaved. The underlying group of the second construction is just (isomorphic to) the quotient [itex]M / IM[/itex]. Furthermore, the two constructions essentially become one-sided inverses -- applying the first construction and then the second construction gives you a module naturally isomorphic to the one you started with.

Roughly speaking, this means that you can view the entire category of R/I-modules as if it was just a subcategory of R-modules. Another way of saying that is that it's unusually simple to use the module theory of R to study the module theory of R/I. How do you identify which R modules are also R/I modules under this correspondence? They are the ones with [itex]I \subseteq Ann(M)[/itex].